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If a bowl contains ten hazelnuts and eight almonds, what is the probability that four nut randomly selected from the bowl will all be hazelnuts?

Answer 1

#0.06862745 ~~6.9%#

We can model the bowl of nuts with a hypergeometric distribution containing 10 "successes" (hazelnuts) and 8 "failures" (almonds).

The formula we want to use is

#P(x=m)=(C_m^M*C_(n-m)^(N-M))/(C_n^N)#

Where:

#N=18# (the population size) #M=10# (the number of successes in the population)

#n=4# (the sample size) #m=4# (the number of successes we want in the sample)

Then the probability of getting 4 hazelnuts in a random sample of 4 nuts is:

#P(x=4)=(C_4^10*C_0^8)/(C_4^18)=(210*1)/(3060)~~0.06862745~~6.9%#

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Answer 2

Julia Adams

The probability of selecting four hazelnuts from the bowl is ( \frac{{\binom{10}{4}}}{{\binom{18}{4}}} ), which simplifies to ( \frac{{210}}{{3060}} ), or approximately 0.0686, or 6.86%.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.